A circuit having a self inductance of 1 henry carries a current of 1 A. To prevent the sparking when the circuit is broken, a capacitor which can withstand 500 V is connected across the switch. What is the minimum value of the capacitance of the capacitor?

Options

A. A. `2 mu F`
B. B. `4 mu F`
C. C. `6 mu F`
D. D. `8 mu F`

Answer

B

Explanation

When the circuit is broken, a very high voltage is developed. Because `e = (d phi)/(dt)` and dt is very very small. The value of C should be such that the energy stored in the capacitor must be equal to the energy stored in the inductor.
`:. 1/2 CV^(2)=1/2 LI^(2)`
`:. C = (LI^2)/(V^2)=(1 xx (1)^(2))/((500)^(2))`
`=(1)/(25 xx 10^(4)) = (10^(-4))/(25) = 4 xx 10^(-6)`
`:. C = 4 mu F`.